Given a cardinal number $|X|$, how many isomorphism classes of schemes with the cardinality of the set of points equal to $|X|$ are there?


closed as off-topic by Steven Landsburg, Dan Petersen, user44191, abx, Todd Trimble Jun 18 at 16:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Steven Landsburg, Dan Petersen, user44191, abx, Todd Trimble
If this question can be reworded to fit the rules in the help center, please edit the question.


For any cardinal $\kappa\neq 0$, there is a proper class of schemes of cardinality $\kappa$ up to isomorphism. (In the case $\kappa = 0$, there is a unique empty scheme up to isomorphism, namely the spectrum of the zero ring.)

Note first that there is a proper class of schemes with just $1$ point. Indeed, there are fields of every infinite cardinality, and for any field $k$, $\text{Spec}(k)$ has only one point. This observation easily extends to the case when $\kappa$ is nonzero and finite, since for any field $k$, $\text{Spec}(k^n)$ has $n$ points.

In the case when $\kappa$ is infinite, we cannot handle arbitrary $\kappa$ with an infinite product of fields. For example, the points of $\text{Spec}(k^\mathbb{N})$ are in bijection with the ultrafilters on $\mathbb{N}$, so $|\text{Spec}(k^\mathbb{N})| = 2^{2^{\aleph_0}}$.

Instead, we can use the fact that $\kappa = \kappa+1$ when $\kappa$ is infinite. Fix an algebraically closed field $F$ of cardinality $\kappa$. For any field $k$, we have $|\text{Spec}(F[x]\times k)| = |\text{Spec}(F[x])\sqcup \text{Spec}(k)| = \kappa$, since $\text{Spec}(F[x])$ has one point for every element of $F$, together with a single generic point, and $\text{Spec}(k)$ has a single point. So again we have a proper class of schemes with underlying set of cardinality $\kappa$.

  • 2
    $\begingroup$ You could also just take the disjoint union of $\kappa$ copies of $\mathrm{Spec}\,k$ (of course your construction is nicer because it produces an affine scheme) $\endgroup$ – Denis Nardin Jun 18 at 14:05